Re: DD simulated by HHH cannot possibly halt (Halting Problem)

On 4/5/2025 5:14 PM, Richard Heathfield wrote:

On 05/04/2025 22:20, olcott wrote:

On 4/5/2025 3:30 PM, Richard Heathfield wrote:

On 05/04/2025 20:21, olcott wrote:

On 4/5/2025 2:12 PM, Richard Heathfield wrote:

On 05/04/2025 20:01, olcott wrote:

As any C programmer can see DDD simulated by HHH would cause
any correct simulator to get stuck in recursive simulation

>
As any C programmer can see, any C compiler is free to reject your code,

>
typedef void (*ptr)();
int HHH(ptr P);
>
void DDD()
{
   HHH(DDD);
   return;
}
>
The above code is correct to the extent that some
HHH is defined somewhere, so it seems that you are
wrong

>
No, you just don't know what I'm talking about. That doesn't surprise me, of course.
>
I'm talking about your HHH code. Any conforming compiler can reject your assembly language inserts.
>

>
I am not taking about that code.

 Yes, you are. And I quote: "As any C programmer can see DDD simulated by HHH"
 

This post has only been about a hypothetical
HHH that simulates its input.

 If you don't wish to be misunderstood, it would be a good idea for you to learn to express yourself more clearly. For example, you could choose better names for your hypothetical functions to avoid confusion between them and functions for which you've published the code on github.
 

I am trying to explain how the actual HHH works
one step at a time. The first step is knowing
how every hypothetical HHH would meet this spec
with DDD as input:
*Simulating termination analyzer Principle*
It is always correct for any simulating termination
analyzer to stop simulating and reject any input that
would otherwise prevent its own termination.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer